Given that the left and right focus of hyperbola are F1 and F2 respectively, P is a point on hyperbola, and Pf1 ⊥ PF2, pf1pf2 = 4AB, then the eccentricity of hyperbola is ■
Let Pf1 be m and PF2 be n
It is defined by hyperbola as: | M-N | = 2A (1)
From the known right triangle pf1f2, there is m ^ 2 + n ^ 2 = (2C) ^ 2 (2)
It is known that Mn = 4AB (3)
If the three equations are simultaneous, then
(1) 2 - (2), get - 2Mn = 4A ^ 2-4c ^ 2 = - 4B ^ 2
That is: B ^ 2 = 2Ab, B = 2A
So B ^ 2 = 4A ^ 2, C ^ 2 = a ^ 2 + B ^ 2 = 5A ^ 2
E = C / a = radical 5
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